CFA Practice Question

There are 434 practice questions for this study session.

CFA Practice Question

The sponsors of a well-known charity came up with a unique idea to attract wealthy patrons to a $500-a-plate dinner. After the dinner, it was announced that each patron attending could buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50-50. If you bought a set of 20 tickets, what is the chance that you will win 15 or more prizes?
A. 0.006
B. 0.250
C. 0.021
Explanation: This is a binomial probability. The probability of getting r successes out of n trials, where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p), is given by: n!(pr)[q(n-r)]/r!(n-r)!. Here, n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore, we have:

P(15) = 20!(0.515)(0.55)/15!5! = 0.0148
P(16) = 20!(0.516)(0.54)/16!4! = 0.0046
P(17) = 0.0011
P(18) = 0.0002
P(19) = 0.00002
P(20) = 0.000001

The sum is 0.0207.

User Contributed Comments 6

User Comment
baddabing Quite easy isn't it!
Xocrevilo The answer provided is too long-winded! You can deduce the answer after just calculating the first term: P(15)=... 0.0148.

0.0148 is greater than 0.006, thus reject answer A.

We also know that the probabilities of P(16), P(17), P(18), P(19) and P(20) will all be smaller than 0.0148, but in sum are unlikely to add up to 0.75, thus reject answers B.
chamad thank you for the tip. For sure the calculation is time consuming in the exam.
aakash1108 .........the problem would begin when the other 3 options are greater than but close to 0.0148.
SCBAnalyst How do you calculate this using a BAII Plus Calculator
at0899 if thats the professional version...
(20 nCr 15 )[(.5)^15][(.5)^5] => .0148
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